Let A = [1 0 0; 2 1 0; 3 2 1]. If u₁ and u₂ are column matrices such that Au₁ = [1; 0; 0] and Au₂ =

Q: Why can't I just find $A^{-1}$ and multiply it by the sum of the result vectors?

You absolutely can, but in the JEE, time is your most precious currency. Finding the inverse of a $3 \times 3$ matrix is computationally expensive and prone to arithmetic errors. Leveraging the linearity property is the 'pro' move.

Q: Is it always true that $A(u_1 + u_2) = Au_1 + Au_2$?

Yes, matrix multiplication is distributive over addition. This is a bedrock property of linear transformations. Whenever you see a sum of vectors being transformed, think of this property first.

Q: Why did we solve the system row by row?

Because $A$ is a lower triangular matrix. In a lower triangular system, the first variable is determined immediately, and each subsequent variable depends only on the ones already found. It's the fastest way to solve a system!

Question

Let $A = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix}$. If $u_1$ and $u_2$ are column matrices such that $Au_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$ and $Au_2 = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}$, then $u_1 + u_2$ is equal to:

Options:

(A) $\begin{bmatrix} -1 \ 1 \ 0 \end{bmatrix}$

(B) $\begin{bmatrix} -1 \ 1 \ -1 \end{bmatrix}$

(C) $\begin{bmatrix} -1 \ -1 \ 0 \end{bmatrix}$

(D) $\begin{bmatrix} 1 \ -1 \ -1 \end{bmatrix}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the world of linear algebra, where the beauty lies not in the brute force of calculation, but in the elegance of the path you choose. Today, we are looking at a matrix

$$A = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix}$$

and two mysterious vectors $u_1$ and $u_2$. The problem asks us to find $u_1 + u_2$.

The amateur approach is to find $u_1$ and $u_2$ individually by inverting $A$ or using Gaussian elimination twice. But you are not an amateur; you are a JEE aspirant. Let's find the smarter way.

### The Trap of Complexity

When you see $Au_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$ and $Au_2 = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}$, your instinct might be to solve for $u_1$ and $u_2$ separately. Resist that urge! It is a trap designed to consume your time.

Instead, look at the structure of the problem. In linear algebra, whenever you see a sum of vectors, your mind should immediately jump to the **distributive property** of matrix multiplication.

### The Power of Linearity

The distributive property states that for any matrix $A$ and vectors $u_1, u_2$, we have $A(u_1 + u_2) = Au_1 + Au_2$. This is the key that unlocks the entire problem.

We don't need to know $u_1$ or $u_2$ individually. We only need to know their sum. Let $u = u_1 + u_2$. Then:

$$Au = Au_1 + Au_2 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} + \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix}$$

Now, the problem has transformed from a complex system into a simple, elegant one:

$$A \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix}$$

### The Final Execution

Look at matrix $A$ again. It is a lower triangular matrix. This is a gift! It means we can solve for $x, y,$ and $z$ using **forward substitution**.

From the first row, $1x + 0y + 0z = 1$, which gives us $x = 1$ instantly.

Moving to the second row, $2x + y = 1$. Substituting $x = 1$, we get $2(1) + y = 1$, so $y = -1$.

Finally, the third row gives $3x + 2y + z = 0$. Substituting $x = 1$ and $y = -1$, we get:

$$3(1) + 2(-1) + z = 0 \Rightarrow 3 - 2 + z = 0 \Rightarrow 1 + z = 0$$

Thus, $z = -1$. Our final vector is:

$$\mathbf{\begin{bmatrix} 1 \ -1 \ -1 \end{bmatrix}}$$

This, my friend, is the essence of the JEE mindset: identify the property, simplify the system, and execute with precision. You have mastered the logic; now go forth and conquer.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 123012001$ . If $u_{1}$ and $u_{2}$ are column matrices such that $A u_{1} = 100$ and $A u_{2} = 010$ , then $u_{1} + u_{2}$ is equal to:

Select Answer:

Visualized Solution

Analyze the Given Matrix $A$

Apply the Linearity Property

Calculate the Resultant Vector

Set up the Matrix Equation

Solve for $x$

Solve for $y$

Solve for $z$

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Trap of Complexity

The Power of Linearity

The Final Execution

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Let $A = [31 - 4 - 1]$ and $B$ be two matrices such that $A^{100} = 100 B + I$ . Then the sum of all the elements of $B^{100}$ is .........

Select Answer:

Visualized Solution

Analyze the Given Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Apply the Linearity Property

Calculate the Resultant Vector

Set up the Matrix Equation

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z